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Example question in a textbook that I don't understand.

Proof works for n = 1

Setting for k makes $k < 2^k $

Setting for k + 1 makes $k+1 < 2^{k+1} $.

Here, I would be stuck, the book takes the equation to:

$k+1<2^k +1\leq 2^k+2^k = 2 \cdot 2^k=2^{k+1}$.

NB: $<2^k +1$ is not a typo.

There doesn't seem to be a good explanation for this in the book (although it does mention adding 1to both sides of the equation), could I have some advice on how the method used works?

user1551
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user100465
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  • Are you wanting to know how the authors even thought about$k+1<2^k +1\leq 2^k+2^k = 2 \cdot 2^k=2^{k+1}$, is that it? – Git Gud Oct 12 '13 at 18:42
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    The book is just adding 1 to both sides of the inductive assumption, that $k < 2^k$. – DanielV Oct 12 '13 at 18:58
  • This problem has been posted several times before, but I'm failing to find duplicates. – Thomas Andrews Oct 12 '13 at 19:01
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    I think everyone here (but me) is misunderstanding the question. As my comment above suggests, I think the OP is trying to understand how one would think about the chain of inequalities at the end of the answer. – Git Gud Oct 12 '13 at 19:02

2 Answers2

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claim : $ n < 2^n$

Proof by Induction theorem.

For $ n = 1 $ . It is true $ 1 < 2^1 = 2 $

Let's say it is true for $ n = k $ $$ k < 2^k \dots (*)$$

Now, if prove it is true for $ n = k+1 $ then it will be true for all k > 1

R.T.P :- $ (k+1)< \mathrm {2}^{k+1},$

we have

$$ k < 2^k \dots (*) $$ Multiply by 2 in (*) , we get

$$ 2k < \mathrm{2}^{k+1} \dots (A) $$

And, we all ready have $ k+1 < 2k, \space \forall k > 1$

Therefore , $ k+1 < 2k < \mathrm {2}^{k+1} $

$ \implies k+1 < \mathrm {2}^{k+1}, $ Hence, proved

Shravan40
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$2^{n+1}=2^n.2>n.2=n+n>n+1$